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D
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Difficulty:
35%
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Question Stats:
70% (01:54) correct
30%
(01:43)
wrong
based on 122
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From a class of 10 students, 5 are to be chosen for a field trip. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the field trip members be chosen?
A. 21
B. 28
C. 35
D. 42
E. 62
A. 21
B. 28
C. 35
D. 42
E. 62
13 Feb 2021, 21:06
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Supposing all 3 will join. Then we have to choose 2 more students from the remaining 7 in 7C2 ways = 7 * 6 / 2 = 21
Supposing all 3 don't joint, then we have to choose 5 from the remaining 7 students in 7C5 = 7C2 = 21
Total number of ways = 21 + 21 = 42
Option D
Arun Kumar
Supposing all 3 don't joint, then we have to choose 5 from the remaining 7 students in 7C5 = 7C2 = 21
Total number of ways = 21 + 21 = 42
Option D
Arun Kumar
15 Feb 2021, 12:35
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Here is my solution for this:
Total # of ways of choosing field trip members = # of ways when all three join + # of ways when all three don't join
# of ways when all three join = There is only one way of selecting all three. The remaining 2 can be picked in 7C2 ways.
# of ways when all three don't join = There is only one way of not selecting all three. The remaining 5 can be picked in 7C5 ways.
Hence Total = (7C2*1)+(7C3*1) = 42 (D)
Total # of ways of choosing field trip members = # of ways when all three join + # of ways when all three don't join
# of ways when all three join = There is only one way of selecting all three. The remaining 2 can be picked in 7C2 ways.
# of ways when all three don't join = There is only one way of not selecting all three. The remaining 5 can be picked in 7C5 ways.
Hence Total = (7C2*1)+(7C3*1) = 42 (D)
